Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 7}{r - 3} \times \dfrac{-5r + 15}{r^2 - 5r - 14} $
First factor the quadratic. $q = \dfrac{r - 7}{r - 3} \times \dfrac{-5r + 15}{(r - 7)(r + 2)} $ Then factor out any other terms. $q = \dfrac{r - 7}{r - 3} \times \dfrac{-5(r - 3)}{(r - 7)(r + 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 7) \times -5(r - 3) } { (r - 3) \times (r - 7)(r + 2) } $ $q = \dfrac{ -5(r - 7)(r - 3)}{ (r - 3)(r - 7)(r + 2)} $ Notice that $(r - 3)$ and $(r - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -5\cancel{(r - 7)}(r - 3)}{ (r - 3)\cancel{(r - 7)}(r + 2)} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $q = \dfrac{ -5\cancel{(r - 7)}\cancel{(r - 3)}}{ \cancel{(r - 3)}\cancel{(r - 7)}(r + 2)} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $q = \dfrac{-5}{r + 2} ; \space r \neq 7 ; \space r \neq 3 $